Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Dr. Wolfgang Sternefeld
Tutoren: Thorwald Kassemek, Natalie Clarius
Semantics 1:
Exercise Sheets
with (partial) solutions
This is a compilation of all exercise sheets and some additional material for the proseminar
Semantics1 in WS 2015/16.
Most (yet not all) of the exercises come with an exemplary solution, which is to be seen as
one possible approach to solve the assignment rather than the one and only gold standard
solution. I do not guarantee absolute and ultimate correctness!
The PDF file contains clickable bookmarks which lead you directly to the desired exercise
sheet.
Have fun!
N.C.
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Dr. Wolfgang Sternefeld
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 1:
Revision
Zimmermann & Sternefeld: Ch. 1-5
Slides ESSLLI: No. 1-3
Gamut: Vol. 1 Ch. 1-4
Partee, Meulen & Wall: Pt. A-B
Hodges: Ch. 1-41
Exercise 1: Statement logic
Proof that the following formulae of propositional logic are equivalent by means of your choice (e.g.
tranformation, truth table, natural deduction, tableaux, ...)
(1) (p ∧ q) ∨ (q → r)
(2) (¬¬p ∧ ¬¬q) ∨ (¬q ∨ r)
(3) ¬ ¬ r ∨ ¬q ∧ ¬ (p ↔ q) ∧ (p ∧ q)
Exercise 2: Predicate logic
Proof that the following formulae of predicate logic are equivalent by means of your choice (e.g. transformation, semantic evaluation, natural deduction, tableaux, ...)
(1) ¬∃x student(x) ∧ ∃y pizza(y) ∧ ¬like(x, y)
(2) ∀x student(x) → ∀y pizza(y) → like(x, y)
(3) ∀x¬ student(x) ∧ ¬∀y¬ pizza(y) ∧ ¬like(x, y)
What do the three of the formulae mean, i.e. expressed in natural langauge?
Exercise 3: Sets
a. Define the set A which consists of all natural numbers, giving the definition in three different ways.
b. Determine the outcome of the following set-theoretic operations:
(1) A ∩ {x ∈ Z : x%2 = 0}1
(2) A \ {x ∈ Z : x is a prime number}
(3) A ∪ {x ∈ Z : x ∈ Z and x < 0}
Hint: You do not need to spell out an infinite number of elements ;) It is sufficient to state enough elements to make the
content of the set clear.
1%
is used here as an operator for retrieving the remainder of a division: 5%3 = 2; 22%7 = 1; 8%4 = 0; ...
– 1/2 –
Exercise 4: Models and interpretations
Let
V = {x, y, z}
be a set of variables and
C = {a, b, c, d}
a set of constants.
Furthermore, assume a model M = hD, Ii with
D = {A, B, C, D}
as denotations of the constants in C.
Specify an interpretation function I such that all of the following formulae are true in that model.
(1) P (a, b)
(2) P (a, d)
(3) R(b, c)
(4) ∀x∃y R(x, y) → Q(y)
(5) ¬∃x∀y¬ R(x, y) → Q(y)
(6) ∀x∃y Q(x) ∧ ∃z R(z, x) → R(x, y)
Exercise 5: Functions and relations
Let A and B be sets such that:
A = {a, b, c}
B = {1, 2, 3}
Furthermore, assume the following sets:
{ha, 1i}, {ha, 2i, hb, 4i}, {ha, ai ha, 1i hc, 3i}, {ha, 1i, hb, 1i, hc, 1i},
{ha, 3i, hb, 1i, hc, 2i}, {ha, 3i, hb, 2i, hc, 3i}, {ha, 2i, ha, 1i, hb, 1i, hc, 3i}
a. Determine which of the above sets are relations R ⊆ A × B and characterise them, that is state if
they are symmetric, reflexive or transitive.
b. Determine which of the above sets are functions f : A → B and characterise them, that is state if
they are injective, surjective or bijective.
Exercise 6: Scope ambiguity and Logical Form
(1) Alle Chinesen haben genau einen einen Kontrabass.
a. For each of the possible readings, rewrite the sentence in predicate logic and specify a model that
makes this reading true but the others false.
b. Briefly explain the ambiguity by using the definition of scope at the level of LF.
– 2/2 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Wolffeld Sternegang
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 2:
Extensions
Zimmermann & Sternefeld: Ch. 4-5
Slides ESSLLI: No. 2-3
Exercise 1: Relational and functional nouns
Assume this little village in Sweden:
Determine the extensions of the following expressions and classify them, i.e. whether they are relational,
functional or of a different category.
(1) neighbour-of
(2) lives-across-from
(3) lives-directly-right-to
(4) has-no-left-neighbour
Solution:
(1) neighbour-of : {< P ippi, M ichel >, < M ichel, P ippi >, < M ichel, Kalle >, ...} ⇒ relational
(2) lives-across-from:{< P ippi, K.u.J. >, < K.u.J., P ipp >, < M ichel, Ronja >, ...} ⇒ functional
(3) lives-directly-right-to: {< P ippi, M ichel >, < M ichel, Kalle >, ...} ⇒ relational
(4) has-no-left-neighbour : {P ippi, K.u.J.} ⇒ neither relational nor functional
– 1/3 –
Exercise 2: Referential expressions
Compute the extensions of the following expressions.
(1) current president of Germany
(2) the current president of Germany
Solution:
(1) ⟦current president of Germany⟧s = {Joachim Gauck}
(2) ⟦the current president of Germany⟧s = Joachim Gauck
Exercise 3: n-place verbs and plugging
a. Let A be a set such that
A = {< a, c >, < a, b >, < c, q >, < d, f >, < a, e >, < y, e >, < x, d >, < q, w >,
< a, y >, < w, e >, < q, c >, < x, a >, < s, w >, < q, a >, < y, x >, < c, d >,
< e, w >, < a, x >, < q, c >, < y, s >, < w, s >, < a, d, s >, < w, s, x >,
< y, s, x >, < a, q, y >, < w, e, d >, < a, w, x >}
Compute the following operations:
→
AÐ
∗s
Ð
A→
∗y
Ð
→
A∗a
Ð
A←
∗q
←
Ð
A∗w
Ð
A←
∗x
(1)
(2)
(3)
(4)
(5)
(6)
b. Given is the following sentence:
(1) John lends his pen to Mary.
1. Calculate the following sets A, B, C, D by use of the plugging operation such that
A.
B.
C.
D.
A = ⟦lends⟧s
B = ⟦lends his pen⟧s
C = ⟦lends his pen to Mary⟧s
D = ⟦John lends his pen to Mary⟧s
2. What is the extension of (1) given that the sentence is false?
Solution:
a. Plugging operations with set A:
(1)
(2)
(3)
(4)
(5)
(6)
→
AÐ
∗ s = {< y >, < w >, < a, d >}
Ð
→
A ∗ y = {< a, q >}
→
AÐ
∗ a = {< x >, < q >}
←
Ð
A ∗ q = {< w >, < c >, < a >}
Ð
A←
∗ w = {< e >, < s >, < s, x >, < e, d >}
←
Ð
A ∗ x = {< d >, < a >, < s >}
– 2/3 –
b. John lends his pen to Mary
1. Plugging operations:
⟦John⟧s = j
⟦his pen⟧s = p
⟦Mary⟧s = m
⟦lends⟧s = {< a, b, c > ∶ a lends c to b}
A. A
= ⟦lends⟧s
= {< j, m, p >, < a, m, p >, < a, b, p >, < a, b, c >}
= R3
B. B
= ⟦lends his pen⟧s
→
= ⟦lends⟧s Ð
∗ ⟦his pen⟧s
Ð
→
=A∗p
→
= {< j, m, p >, < a, m, p >, < a, b, p >, < a, b, c >}Ð
∗p
= {< x, y > ∣ < p >∈ {< j, m, p >, < a, m, p >, < a, b, p >, < a, b, c >}}
= {< j, m >, < a, m >, < a, b >}
= R2
C. C
= ⟦lends his pen to Mary⟧s
→
= ⟦lends his pen⟧s Ð
∗ ⟦Mary⟧s
Ð
→
=B∗m
→
= {< j, m >, < a, m >, < a, b >}Ð
∗m
= {< x > ∣ < m >∈ {< j, m >, < a, m >, < a, b >}}
= {< j >, < a >}
= {j, a}
= R1
D. D
= ⟦John lends his pen to Mary⟧s
→
= ⟦John⟧s Ð
∗ ⟦lends his pen to Mary⟧s
Ð
→
=j∗C
→
= CÐ
∗j
→
= {< j >, < a >}Ð
∗j
= {<> ∣ < j >∈ {< j >, < a >}}
= {<>}
= R0
=1
2. If the sentence is false, then its extension is ∅.
– 3/3 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Wolfgang Feldstern
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 3:
Extensions, Determiners and Quantifiers
Zimmermann & Sternefeld: Ch. 5-6
Slides ESSLLI: No. 3-4 + Addenda
Exercise 1: Plugging
Given are the following expressions:
⟦Paul ⟧s = p
⟦sleeps⟧s = {< p >, < a >}
Compute the plugging operation
⟦sleeps⟧s ∗ ⟦p⟧s
and determine its meaning in natural language.
Solution:
⟦sleeps⟧s ∗ ⟦p⟧s
= {< p >, < a >} ∗ p
= {∅}
Meaning in natural language: Paul is sleeping
Exercise 2: Predicate modification
Given are the following sets:
⟦female⟧s = {M arge, M aggie, Sharon, Lisa}
⟦male⟧s = {Kyle, Eric, Randy, Kenney, Sharon, Bart, Homer, Stan}
⟦child ⟧s = {Kyle, Eric, Bart, Stan, Lisa, M aggie, Kenney}
⟦adult⟧s = {Randy, Sharon, M arge, Homer}
⟦from⟧s = {< x, y > ∣ x is f rom y} =
{< Kyle, Southpark >, < Eric, Southpark >, < Randy, Southpark >, < M arge, Springf ield >,
< Bart, Springf ield >, < Stan, Southpark >, < Homer, Springf ield >, < Kenney, Southpark >,
< M aggie, Springf ield >, < Lisa, Springf ield >, < Sharon, Southpark >}
Compute
a. ⟦the [female [adult from Southpark]] ⟧s
b. ⟦the [[female adult] [from Southpark]] ⟧s
and explain the similarity or difference between the results.
Solution:
The results will be identical since set intersection is associative: A ∩ (B ∩ C) = (A ∩ B) ∩ C
– 1/2 –
Exercise 3: Determiners and Quantifiers
Calculate the truth conditions for the sentence
⟦no student reads the book ⟧s
according to the definitions in chapter 6.1 of the book.
Solution:
⟦no student reads the book ⟧s = 1
iff ⟦reads the book ⟧s ∈ ⟦no student⟧s
Ð
iff ⟦reads the book ⟧s ∈ (⟦no⟧s ←
∗ ⟦student⟧s )
Ð
iff ⟦reads the book ⟧s ∈ ({< X, Y > ∣ X ∩ Y = ∅}←
∗ {x ∣ x is a student in s})
iff ⟦reads the book ⟧s ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
→
iff (⟦reads⟧s Ð
∗ ⟦the book ⟧s ) ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
Ð
→
iff (⟦reads⟧s ∗ (⟦the⟧s (⟦book ⟧s ))) ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
→
iff (⟦reads⟧s Ð
∗ (⟦the⟧s ({x ∣ x is a book in s}))) ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
Ð
→
iff (⟦reads⟧s ∗ ({< X, y > ∣ X = {y}}({x ∣ x is a book in s}))) ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
→
iff (⟦reads⟧s Ð
∗ thebook) ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
→
iff ({< x, y > ∣ x reads y in s}Ð
∗ thebook) ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
iff {x ∣ x reads the book in s} ∈ {Y ∣ {x ∣ x is a student in s} ∩ Y = ∅}
iff {x ∣ x is a student in s} ∩ {x ∣ x reads the book in s} = ∅
– 2/2 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Dr. Wolfgang Sternefeld
Tutoren: Thorwald Kassemek, Natalie Clarius
Exemplary Solution:
QDPs in Object Position
Zimmermann & Sternefeld: Ch. 6
Slides ESSLLI: No. 4 + Addenda
The sentence
Ein Bild
zierte
jede Wand.
A
picture decorated every wall.
’A picture decorated every wall.’
Our sentence to be evaluated has two readings:
1. the linear reading
ˆ with ein Bild having wide scope w.r.t. jede Wand :
ˆ There is at least one picture for which it holds that that picutre decorated every wall
ˆ ∃x(Bild(x) ∧ ∀y(W and(y) → Zierte(x, y)))
ˆ
S
VP
QDP
ein Bild
QDP
V
zierte jede Wand
2. the inverted reading
ˆ with jede Wand having wide scope w.r.t. ein Bild :
ˆ For every wall it holds that there is at least one picture that decorated that wall
ˆ ∀y(W and(y) → ∃x(Bild(x) ∧ Zierte(x, y)))
ˆ
S
QDP
jede Wandy
{y ∶ ein Bild zierte ty }
S
y:
VP
QDP
ein Bild
V
DP
zierte
ty
You do not need to worry about the precise syntactic labels here; the tree is just there to show in
what way the sentence is composed out of the types of the individual expressions.
– 1/6 –
Intuitively, reading (1) should be false and reading (2) should be true. The task is to formally proof this
intuition by calculation of truth conditions.
The model
Modified model M =< D, I >:
D = {A, B, C, D, a, b, c, d}
I(Bild) = {a, b, c, d}
I(W and) = {A, B, C, D}
I(zierte) = {< a, A >, < b, B >, < c, C >< d, D >}
Note that this model now contains the additional picture d (included as the constant d in the domain D
as well as in the interpretation of Bild ) decorating the wall D (included as the relation < d, D > in the
interpretation of zierte).
– 2/6 –
In situ interpretation
will give us the linear reading.
⟦Ein Bild zierte jede Wand ⟧s
= ⟦Ein Bild ⟧s ∗Q ⟦zierte jede Wand ⟧s
The first step is to higher-order-plug the QDP ⟦ein Bild ⟧s with the VP, meaning that the QDP is meant
to fill out the subject position of the VP. We do not need an arrow here since the VP in which the object
is already included leaves only one position open (namely the subject position: ⟦zierte jede Wand ⟧s =
{x ∶ x decorated every wall in s}) so we don’t have to specify from what side we are plugging. The Q
is the indicator for plugging with a set of sets.
Note that at this point, it is not obligatory yet to apply higher-order plugging: We could as well apply the
default plugging operation, meaning that the VP has to be an element of the QDP. This does, however,
not work with objects that are QDPs, here we do have to make use of higher-order plugging. But it does
not harm to also do so for the subject, if we do so, the sentence after plugging is
= {<> ∶ {x ∶ x ∈ ⟦zierte jede Wand ⟧s } ∈ ⟦ein Bild ⟧s }
which is the set of all 0-tuples such that the set of things that decorated every wall is an element
of ⟦ein Bild ⟧s , which is equal to the stamement
= 1 iff ⟦zierte jede Wand ⟧s ∈ ⟦ein Bild ⟧s
meaning that the sentence turns out to be true iff the VP (a set of individuals, namely those that
decorated every wall) is an element of the QDP (a set of sets, namely those predicates that apply to a
picutre).
Next we have to further decompose the VP into the verb itself and the object. Since ⟦jede Wand ⟧s
is a QDP as well, we have to apply
= 1 iff (⟦zierte⟧s Ð
∗→
Q ⟦jede Wand ⟧s ) ∈ ⟦ein Bild ⟧s
higher-order plugging between the verb ⟦zierte⟧s and the QDP object ⟦jede Wand ⟧s . Here we do need
the arrow: The relation ⟦zierte⟧s now stands alone and has two argument positions open (⟦zierte⟧s =
{< x, y > ∶ x decorated y in s}) of which ⟦jede Wand ⟧s is supposed to fill the object position; therefore,
we need to indicate that we plug at the right side of the relation (y).
What we get now is the following:
= 1 iff {x ∶ {y ∶ < x, y >∈ ⟦zierte⟧s } ∈ ⟦jede Wand ⟧s } ∈ ⟦ein Bild ⟧s
= 1 iff {x ∶ {y ∶ x decorated y in s} ∈ ⟦jede Wand ⟧s } ∈ ⟦ein Bild ⟧s
What we have here is the set of all subjects for which it holds that the set of all objects which are
in the decorating-relation with the subject must be an element of ⟦jede Wand ⟧s , and this set of subjects
that are in the decorating-relation with every wall must again be an element of ⟦ein Bild ⟧s .
Now we have to check what is in contained in this set. We have to go through every possible value
for x (for all decorators, i.e. the pictures), check what are in the inner set the values for y (the decorated,
i.e. the walls) and whether they are in ⟦jede Wand ⟧s , and for those y that fulfil the condition we can
put the corresponding x into our outer set, and this set must again be an element of ⟦ein Bild ⟧s . So let’s
assign values vor x (the decorators) and then check what it is that they decorate:
For x = a ∶
⟦a zierte jede Wand ⟧s
= 1 iff {y ∶ a decorated y in s} ∈ ⟦every wall ⟧s
= 1 iff {A} ∈ ⟦every wall ⟧s
= false
Therefore, a ∉ {x ∶ {y ∶ < x, y >∈ ⟦zierte⟧s ∈ ⟦jede Wand ⟧s }}
Therefore, a ∉ {x ∶ {y ∶ x decorated y in s} ∈ ⟦jede Wand ⟧s } ∈ ⟦ein Bild ⟧s
– 3/6 –
We see that the things that are decorated by the picture a is not in ⟦every wall ⟧s , therefore the picture
a is not contained in the set of things that decorated every wall.
We get a similar result for all other picutres:
For x = b ∶
⟦b zierte jede Wand ⟧s
= 1 iff {y ∶ b decorated y in s} ∈ ⟦every wall ⟧s
= 1 iff {B} ∈ ⟦every wall ⟧s
= false
Therefore, b ∉ {x ∶ {y ∶ < x, y >∈ ⟦zierte⟧s ∈ ⟦jede Wand ⟧s }}
Therefore, b ∉ {x ∶ {y ∶ x decorated y in s} ∈ ⟦jede Wand ⟧s } ∈ ⟦ein Bild ⟧s
... and the same for the picutres c and d.
So the set of pictures that decorated every wall is the empty set:
{x ∶ {y ∶ < x, y >∈ ⟦zierte⟧s ∈ ⟦jede Wand ⟧s }}
= {x ∶ {y ∶ x decorated y in s} ∈ ⟦jede Wand ⟧s } ∈ ⟦ein Bild ⟧s
=∅
Now in order to make the stament true, this set must be an element of ⟦ein Bild ⟧s , namely the set
of all sets that share at least one element with ⟦Bild ⟧s :
= 1 iff ∅ ∈ ⟦ein Bild ⟧s
= 1 iff ∅ ∈ {Y ∶ ⟦Bild ⟧s ∩ Y ≠ ∅}
= 1 iff ∅ ∩ {a, b, c, d} ≠ ∅
= f alse
We see that there is no element that decorated every wall that is also a picture, meaning that there
is not at least one picture that decorated every wall. Therefore we get that the linear reading is false –
which is precisely what we wanted.
– 4/6 –
Quantifier raising
will give us the inverted reading.
⟦Ein Bild zierte jede Wand ⟧s
= ⟦jede Wand y ⟧s ∗ ⟦Ein Bild zierte ty ⟧s
Moving the QDP out of the sentence results in set formation for the sentence it has been moved out,
and the plugging operation turns out to be true iff the set of things that fulfil the VP of being decorated
by a picture is an element of the QDP:
= 1 iff ⟦ein Bild zierte ty ⟧s ∈ ⟦jede Wand ⟧s
= 1 iff {y ∶ ⟦ein Bild zierte y⟧s = 1} ∈ ⟦jede Wand ⟧s
Now we again have to decompose the VP into the two-place relation and the subject:
= 1 iff {y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s } ∈ ⟦ein Bild ⟧s } ∈ ⟦jede Wand ⟧s
= 1 iff {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s } ∈ ⟦jede Wand ⟧s
Here we have a very similar syntax to what we already got when applying higher-order plugging above,
just the other way round: Here we have the set of objects such that the set of subjects that decorated it
are an element of ⟦ein Bild ⟧s , and this set of objects that are decorated by a picture again has to be an
element of ⟦every wall ⟧s .
Now we again have to assign values for y (the decorated, i.e. the walls) in the outer set, check what
are in the inner set the x that decorated it (the decorators, i.e. the pictures), whether they fulfil the
property of being in ⟦ein Bild ⟧s and if so, whether the set of walls that are decorated by a picture is in
the semantics of ⟦jede Wand ⟧s . So let’s assign values for y:
For y = A:
⟦ein Bild ziert A⟧s
= 1 iff {x ∶ x decorated A in s} ∈ ⟦ein Bild ⟧s
= 1 iff {a} ∈ ⟦ein Bild ⟧s
= 1 iff {a} ∩ {a, b, c, d} ≠ ∅
= true
Therefore, A ∈ {y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s } ∈ ⟦ein Bild ⟧s } ∈ ⟦jede Wand ⟧s
Therefore, A ∈ {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s }
So the wall A belongs to those walls that are decorated by at least one picture.
The same calculation holds for the other walls:
...
For y = D:
⟦ein Bild ziert D⟧s
= 1 iff {x ∶ x decorated D in s} ∈ ⟦ein Bild ⟧s
= 1 iff {d} ∈ ⟦ein Bild ⟧s
= 1 iff {d} ∩ {a, b, c, d} ≠ ∅
= true
Therefore, D ∈ {y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s } ∈ ⟦ein Bild ⟧s } ∈ ⟦jede Wand ⟧s
Therefore, D ∈ {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s }
Hence, the set of walls that are decorated by a picutre consists of all the walls A,B,C and D:
{y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s } ∈ ⟦ein Bild ⟧s } ∈ ⟦jede Wand ⟧s
= {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s }
= {A, B, C, D}
and this set has to be in ⟦every wall ⟧s , it must be an element of those predicates that apply to ev-
– 5/6 –
ery wall (and therefore the set of walls must be a subset of the elements being decorated by a picture):
= 1 iff {A, B, C, D} ∈ ⟦jede Wand ⟧s = 1 iff {A, B, C, D} ∈ {Y ∶ ⟦Wand ⟧s ⊆ Y }
= 1 iff {A, B, C, D} ∈ {A, B, C, D}
= true
We see that in fact, every wall fulfils the property of being decorated by a picture. Hene, we see
that the inverted reading, in which it must hold that for every wall there must be at least one picture
which decorated that wall, is true – which is again exactly what we wanted.
– 6/6 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Dr. Wolfstern Feldgang
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 4:
Types
Zimmermann & Sternefeld: Ch. 6
Gamut: Vol. 2 Ch. 4
Exercise 1: Types of expressions
Determine the logical types of the following expressions using trees:
a. eat
b. the chocolate
c. eat the chocolate
d. Klaus eats the chocolate
e. Klaus eats the chocolate he bought
Solution:
Klaus
eats
the
chocolate Ch{x : he bought x }
<et>
<et>1
<<et>e>
<e<et>>
e
<et>
e
<et>
t
1
This is an exceptional case where we do NOT have functional application of types but predicate
modification, i.e. intersection of two sets yielding a new set (or, more precisely, its characteristic funtion
of type <et>).
– 1/4 –
Alternative:
Klaus
eats
the
chocolate
P redM od
(Ch{x : he bought x })
<<<et><<et><et>>>>
<et>
<et>
<<et>e>
<e<et>>
e
<<et><et>>
<et>*
e
<et>
t
ˆ Operator Op wird (wie bei QR) auf LF bewegt: Opi he bought ti , mit he bought ti vom Typ <t>
ˆ aus dem Satz wird (wie bei QR) die Menge gebildet: Ch{x: he bought x } vom Typ <et>
ˆ Op selbst (also Verküpfung zwischen NP und RS in Form von PredMod) fungiert als eine Art
Identitätsfunktion vom Typ <<et><et>>
ˆ Semantik des PredMod-Operators: Für alle P, Q vom Typ <et> ist Op(P )(Q) (d.h. die Anwendung des Operators wird in zwei Teilschritte zergliedert und zuerst auf den RS angwendet, dann
auf die NP!) diejenige Funktion F vom Typ <et>, sodass für alle x vom Typ <e> gilt: F (x) = 1
gdw P (x) und Q(x) = 1.
ˆ erster Schirtt: PredMod-Operator vom Typ <<<et><<et><et>>>> nimmt die Menge vom
Typ <et> als Argument und gibt als Wert die Identitätsfunktion vom Typ <<et><et>> aus, die
durch Funktionsanwendung auf chocolate wieder eine Eigenschaft macht
ˆ zweiter Schritt: Anwendung der (den RS bereits enthaltenden) Identitätsfunktion vom Typ <<et><et>>
auf chocolate vom Typ <et> ergibt die gePredModte NP vom Typ <et>
– 2/4 –
Exercise 2: Composition of types
Explain the problem with compositional type assignment in the sentence
Klaus eats a chocolate
and one way this problem can be resolved.
Problem:
eats
Klaus
a
chocolate
<<et><<et>t>>
<et>
<e<et>>
e
<<et>t>
?? Type mismatch
??
Solution 1:
In situ (type shifting ”eats”)
Klaus
eats
a
chocolate
<e<et>>
<<et><<et>t>>
<et>
<<<et>t><et>>
e
<<et>t>
<et>
t
ˆ Type-Shift des Verbs:
● Lif t(V für Indiv-Obj ) = V für QDP-Obj
● Lif t(<e<et>>) = <<<et>t><et>>
● zweist. Verb mit Indiv-Objekt als Argument (und Menge von Subjekten als Wert)
wird angehoben zu
zweist. Verb mit QDP-Objekt als Argument (und Menge von Subjekten als Wert)
ˆ Semantik des Lift-Operators:
● Altes Format (ohne Typen): Lif t(R)(QDP ) = {x ∶ {y ∶ < x, y > ∈ R} ∈ QDP }
● Neues Format (mit Typen): Lif t(R)(QDP ) = Ch{x ∶ QDP (Ch{y ∶ R(y)(x) = 1}) = 1}
mit R vom Typ <e<et>> und QDP vom Typ <<et>t>
– 3/4 –
Solution 2:
QR (raising ”a chocolate”)
S
S’
QDP
a chocolatey
VP
DP
V
QDP
eats
ty
Klaus
a
{y:
chocolatey
Klaus
eats
ty }
<e<et>> <e>
<et>
e
<<et><<et>t>>
<et>
<t<et>>
<<et>t>
<t>
<et>
t
– 4/4 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Feldgang Wolfstern
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 5:
Propositions and intensions
Zimmermann & Sternefeld: Ch. 7-8
Slides ESSLLI: No. 5
Exercise 1: Propositions
Consider the following worlds:
w1
w2
w3
Determine
a. ⟦John loves Mary⟧
b. ⟦Mary loves John⟧
c. ⟦John loves Mary⟧w for each w
d. ⟦Mary loves John⟧w for each w
What are the propositions expressed by the sentences?
Solution:
a. ⟦John loves Mary⟧
= {w ∶ ⟦John loves Mary⟧w = 1}
= {w1 , w2 }
b. ⟦Mary loves John⟧
= {w ∶ ⟦Mary loves John⟧w = 1}
= {w1 , w3 }
c. ⟦John loves Mary⟧w1
= 1 iff w1 ∈ ⟦John loves Mary⟧
=1
d. ⟦John loves Mary⟧w2
= 1 iff w2 ∈ ⟦John loves Mary⟧
=1
e. ⟦John loves Mary⟧w3
= 1 iff w3 ∈ ⟦John loves Mary⟧
=0
– 1/3 –
w4
f. ⟦John loves Mary⟧w4
= 1 iff w4 ∈ ⟦John loves Mary⟧
=0
g. ⟦Mary loves John⟧w1
= 1 iff w1 ∈ ⟦Mary loves John⟧
=1
h. ⟦Mary loves John⟧w2
= 1 iff w2 ∈ ⟦Mary loves John⟧
=0
i. ⟦Mary loves John⟧w3
= 1 iff w3 ∈ ⟦Mary loves John⟧
=1
j. ⟦Mary loves John⟧w4
= 1 iff w4 ∈ ⟦Mary loves John⟧
=0
The propositions of the sentences are the sets of posssible worlds of which that sentence is true.
– 2/3 –
Exercise 2: Intensional constructions
Determine the truth conditions for the sentences
a. ⟦Günther knows that Hildegard sleeps⟧w
b. ⟦Gisela is a former US president⟧w
c. ⟦Hans-Peter searches a unicorn⟧w
Solution:
a. ⟦Günther knows that Hildegard sleeps⟧w
= ⟦subject + attitude verb + that + S ⟧w
→
= ⟦Günther ⟧w ∗ (⟦knows⟧w Ð
∗ ⟦Hildgerad sleeps⟧)
→
= ⟦Günther ⟧w ∗ ({< x, S > ∶ < x, S >∈ ⟦knows⟧w }Ð
∗ ⟦Hildgerad sleeps⟧)
= ⟦Günther ⟧w ∗ {x ∶ < x, ⟦Hildgerad sleeps⟧ >∈ ⟦knows⟧w }
= 1 iff < ⟦Günther ⟧w , ⟦Hildgegard sleeps⟧ >∈ ⟦know ⟧w
= 1 iff Günther knows that Hildgerad sleeps in w
b. ⟦Gisela is a former US president⟧w
= ⟦subject + intensional-adjective + noun⟧w
= ⟦Gisela⟧w ∗ (⟦former ⟧w (⟦US president⟧))
= ⟦Gisela⟧w ∗ ⟦former US president⟧w
= ⟦Giesla⟧w ∗ {x ∶ x is a former US president in w}
= 1 iff ⟦Gisela⟧w ∈ {x ∶ x is a former US president in w}
= 1 iff Gisela is a former US president in w
c. ⟦Hans-Peter searches a unicorn⟧w
= ⟦subject + opaque verb + QDP ⟧w
(a) De dicto:
= ⟦Hans-Peter ⟧w ∗ (⟦searches⟧w Ð
∗→
Q ⟦a unicorn⟧)
= ⟦Hans-Peter ⟧w ∗ {x ∶ {y ∶ x searches y} ∈ ⟦a unicorn⟧}
= ⟦Hans-Peter ⟧w ∗ {x ∶ x searches a unicorn in w}
= 1 iff ⟦Hans-Peter ⟧w ∈ {x ∶ x searches a unicorn in w}
= 1 iff Hans-Peter searches a unicorn in w
(b) De re:
= ⟦a unicorn⟧ ∗ {y ∶ (⟦Hans-Peter ⟧w ∗ {x ∶ x searches a unicorn in y}) = 1}
= ⟦a unicorn⟧ ∗ {y ∶ Hans-Peter searches y in w}
= 1 iff {y ∶ Hans-Peter searches y in w} ∈ ⟦a unicorn⟧
= 1 iff there is a unicorn that Hans-Peter searches in w
– 3/3 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Gangstern Feldwolf
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 6:
Presuppositions I
Zimmermann & Sternefeld: Ch. 9
Levinson: Ch. 4
Exercise 1: Presupposed statements
For each of the sentences below, state all (obvious) presuppositions this sentence has (in case the sentence
does have one at all) and if possible, determine the exact nature of the presupposition (e.g. existence
presupposition) and its trigger (e.g. a determiner, a factive verb, ...):
(1) John’s wife is pretty
(2) The neighbour of John has a dog
(3) John didn’t know that Mary is married
(4) John doesn’t believe that elephants are not pink
(5) John proved that he is innocent
(6) John has given up playing tennis
Solution:
(2) The neighbour of John has a dog
There is a neighbour of John (existence prsp.)
There is only one neighbour of John (uniqueness prsp.)
(1) John’s wife is pretty
John has a wife (existence prsp., triggered by the definite description)
both triggered by the
(4) John doesn’t believe that elephants are not pink
no prsp.s because believe is not a veridical verb
(5) John proved that he is innocent
John is innocent (manage is a veridical, but not a factive verb; the prsp. doesn’t hold for the
(3) John didn’t know that Mary is married
Marry is married (know is a veridical and a factive verb)
negated sentence)
(6) John has given up playing tennis
John has played tennis in the past (give up is a change-of-state-verb requiring a complement
sentence which was true at some point in the past)
1
Exercise 2: Truth of statements and presuppositions
a. Given the sentence
S1: The queen of Antarctica has green hair
give the presupposition(s) of S1 and determine its extension in case the presupposition is true and
in case it is false, each of the cases once according to Frege’s naive analysis and once according
Russell’s quantificational analysis (that means you have to give 4 extensions in total). Briefly
explain the difference.
b. Now consider the sentence
S2: The queen of Antarctica doesn’t have green hair
Which presuppositions does the negated sentence entail in each of the two analyses, and why?
Solution:
a. S1: The queen of Antarctica has green hair
(a) Presuppositions:
P: There is exactly one queen of Antarctica, consisting of
P-a: There is at least one queen of Antarctica (existence condition)
P-b: There is no more than one queen of Antarctica (uniqueness condition)
(b) Extensions:
⟦S1 ⟧w with Frege
⟦S1 ⟧w with Russell
⟦P ⟧w =1
1 iff the QoA has green hair, 0 else
1 iff the QoA has green hair, 0 else
⟦P ⟧w = 0
undefined (truth value gap)
0
b. S2: The queen of Antarctica doesn’t have green hair
(a) Frege: S2 entails P:
The QoA doesn’t have green hair
⊧There is exactly one QoA
(b) Russell:
S2 = logical negation of S1 (wide scope) does NOT entail P:
¬∃x(Queen(x) ∧ ¬∃y((x ≠ y) ∧ Queen(y)) ∧ Greenhair(x))
⊧
/ ∃x(Queen(x) ∧ ¬∃y((x ≠ y) ∧ Queen(y)))
only negation of the predicate (narrow scope) does:
∃x(Queen(x) ∧ ¬∃y((x ≠ y) ∧ Queen(y)) ∧ ¬Greenhair(x))
⊧ ∃x(Queen(x) ∧ ¬∃y((x ≠ y) ∧ Queen(y)))
2
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Feldwolf Sternegang
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 7:
Presuppositions II
Zimmermann & Sternefeld: Ch. 9
Exercise 1: Presupposition and assertion
For each of the sentences below, state its presupposition and its assertion:
(1) a. John has eaten all the chocolate again
b. John will never eat all the chocolate again
(2) a. John stops annoying Mary
b. John continues annoying Mary
(3) a. John ’s cat likes mice
b. John hasn’t succeeded in installing the lamp
(4) a. John knows that the shop closes at 20.00h
b. John regrets that he missed his favourite show
Solution:
(2) a. John stops annoying Mary
P: John has been annoying Mary in the past
A: John doesn’t and won’t annoy Mary
b. John continues annoying Mary
P: John has been annoying Mary in the past
A: John does and will annoy Mary
(3) a. John ’s cat likes mice
P: John has a cat
A: It likes mice
b. John hasn’t succeeded in installing the lamp
P: John tried to install the lamp
A: John hasn’t succeeded in installing the lamp
(4) a. John knows that the shop closes at 20.00h
P: The shop closes at 20.00h
A: John knows that the shop closes at 20.00h
b. John regrets that he missed his favourite show
A: John missed his favourite show
P: John regrets that he missed his favourite show
(1) a. John has eaten all the chocolate again
P: John ate all the chocolate at some points in the past
A: John has eaten all the chocolate
b. John won’t eat all the chocolate again
P: John ate all the chocolate at some points in the past
A: John won’t eat all the chocolate
– 1/4 –
Exercise 2: Context change
Assume the common ground at the beginning of a conversation
CG = {w1 , w2 , w3 , w4 , w5 , w6 , w7 , w8 , w9 }
and the the propositions of the sentences
⟦S1 ⟧ = {w2 , w4 , w5 , w6 , w7 }
⟦S2 ⟧ = {w1 , w2 , w4 , w6 , w9 }
⟦S3 ⟧ = {w1 , w2 , w3 , w4 , w5 , w6 , w7 , w8 , w9 }
⟦S4 ⟧ = {w1 , w2 , w5 , w7 , w9 }
Calculate the updated CG after each of the uttereances as well as their informativity:
1. S1 : John is drinking fresh apple juice from his gramdother
2. S2 : John is at home
3. S3 : Either roses are red or they are not red
4. S4 : The sun is shining
Solution:
1. CG′ = CG ∩ ⟦S1 ⟧
= {w1 , w2 , w3 , w4 , w5 , w6 , w7 , w8 , w9 } ∩ {w2 , w4 , w5 , w6 , w7 }
= {w2 , w4 , w5 , w6 , w7 }
informativity = ∣CG∣ − ∣CG′ ∣ = 9 − 5 = 4
2. CG′′ = CG′ ∩ ⟦S2 ⟧
= {w2 , w4 , w5 , w6 , w7 } ∩ {w1 , w2 , w4 , w6 , w9 }
= {w2 , w4 , w6 }
informativity = ∣CG′ ∣ − ∣CG′′ ∣ = 5 − 3 = 2
3. CG′′′ = CG′′ ∩ ⟦S3 ⟧
= {w2 , w4 , w6 } ∩ {w1 , w2 , w3, w4 , w5 , w6 , w7 , w8 , w9 }
= {w2 , w4 , w6 }
informativity = ∣CG′′ ∣ − ∣CG′′′ ∣ = 3 − 3 = 0
4. CG′′′′ = CG′′′ ∩ ⟦S4 ⟧
= {w2 , w4 , w6 } ∩ {w1 , w2 , w5 , w7 , w9 }
= {w2 }
informativity = ∣CG′′′ ∣ − ∣CG′′′′ ∣ = 3 − 1 = 2
– 2/4 –
Exercise 3: Accomodation
Assume that John successfully utters
S1 : My wife is sick
and Mary didn’t know before that
S2 : John has a wife
Furthermore, assume
W = {w1 , w2 , w3 , w4 , w5 , w6 , w7 , w8 }
⟦S2 ⟧ = {w1 , w2 , w3 , w4 }
⟦S1 ⟧ = {w1 , w2 }
Determine each state of Mary’s version of the CG and draw the corresponding Euler diagrams
a. before S1 was uttered
b. after the accomodation but before the CG update
c. after the CG update
Solution:
a. Initial CG:
CG = W
= {w1 , w2 , w3 , w4 , w5 , w6 , w7 , w8 }
b. Accomodation to S2 :
CG′ = CG ∩ ⟦S2⟧
= {w1 , w2 , w3 , w4 , w5 , w6 , w7 , w8 } ∩ {w1 , w2 , w3 , w4 }
= {w1 , w2 , w3 , w4 }
c. CG update to S1 :
CG′′ = CG′ ∩ ⟦S1⟧
= {w1 , w2 , w3 , w4 } ∩ {w1 , w2 }
= {w1 , w2 }
– 3/4 –
Exercise 4: Presupposition projection
Given:
S: If John has a car, his car is broken
S(a): John has a car
S(b): His car is broken
a. Determine the modified CG and draw a corresponding Venn diagram according to the definitions
in the book.
b. Determine the presupppositions of each sentence.
c. State which presuppositions project and which do not, and explain why.
Solution:
a. Euler diagram:
CG′ = CG ∖ ((CG ∩ S(a) ∖ ((CG + S(a)) ∩ S(b)))))
= CG - (rechte Hälfte - (oberes rechtes Viertel))
= CG - unteres rechtes Viertel
= CG - (S(a) ∧ ¬S(b))
= CG + S(a) → S(b)
b. S(b) presupposes S(a) (i.e. is a subset of ⟦S(a)⟧; undefined for ⟦not S(a)⟧)
c. The presupposition S(a) does not project but is blocked by the if-else-construction producing
CG′ = CG ∖ (S(a) ∩ not S(b)); therfore the whole sentence S does not have a presupposition (i.e.
⟦S⟧ ⊆/ ⟦S(a)⟧ and ⟦S⟧ ⊆/ ⟦not S(a)⟧)
– 4/4 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Gangfeld Sternewolf
Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 8:
Predicate Logic
Zimmermann & Sternefeld: Ch. 10
Gamut: Vol. 1 Ch. 1-4
Gamut: Vol. 2 Ch. 4-5
Partee, Meulen & Wall: Pt. A-B, D
Hodges: Ch. 1-41
Exercise 1: Predicate logic
Translate the following statments into predicate logic; in case there are two readings, state both of them:
(1) All men have a dream
(2) John loves Mary iff Mary loves John
(3) All children like Harry Potter
(4) John owns all Harry Potter books
(5) John likes the car he bought and Peter does too
(6) John and Mary buy a house
Solution:
(1) ∀x(man(x) → ∃y(dream(y) ∧ have(x, y)))
OR
∃x(dream(x) ∧ ∀y(man(y) → have(y, x)))
(2) love(j, m) ↔ love(m, j)
(3) ∀x(chilren(x) → like(x, hp))
(4) ∀x(hpb(x) → own(j, x))
(5) Strict reading:
∃x(((car(x) ∧ bought(j, x)) ∧ (like(j, x) ∧ like(p, x))) ∧ ∀y((car(y) ∧ bought(j, y)) → (x = y)))
OR
Sloppy reading:
⎛
⎞
∃x (((car(x) ∧ bought(j, x)) ∧ like(j, x))) ∧ ∀y((car(y) ∧ bought(j, y)) → (x = y)))) ∧ ...
⎝
⎠
⎛
⎞
... ∃z ((car(z) ∧ bought(p, z)) ∧ like(p, y))) ∧ ∀a((car(a) ∧ bought(p, a)) → (z = a))
⎝
⎠
(6) ∃x(house(x) ∧ (buy(j + m, x)))
OR
∃x(house(x) ∧ buy(j + m, x)) ∧ ∃y(house(y) ∧ buy(m, y))
– 1/5 –
Exercise 2: Variable assignments
Given the model
M =< D, I >
D = {John, M ary, P eter}
I(man) = {John, P eter}
I(love) = {< John, M ary >, < P eter, M ary >, < M ary, P eter >}
formalize and evaluate the sentence
There is a man who loves Mary
using a derivation tree.
Solution:
(Ähnlich wie der Baum in MathMethLing S. 328, für jeden Knoten die Denotation unter M, g, d.h.
⟦x⟧M,g = F (x) = g(x) = M ary etc.
Eigentlich wird wegen der Auswertung des Existenzquantors für jedes g ein eigener Baum benötigt, der
die Denotation des Satzes unter der jeweiligen Belegung darstellt; der Baum rechts ist exemplarisch für
den Fall g(x) = M ary.)
0
man(x) ∧ love(x, m)
man(x)
man x
love(x, m)
love x m
0
0
{John, P eter} M ary {< J., M. >, ...} M ary M ary
Semantics of the existential quantifier:
There is an assignment function g s.t. the proposition (man(x) ∧ (love(x, m)) is true under g (with
g(x) = John or g(x) = P eter), thefore the whole statement is true.
– 2/5 –
Exercise 3: Lambda expressions
Formalize the following expressions using the lambda calculus step by step (i.e. compute a suitable
lambda expression for every constituent and compositionally apply them to each other), using names for
predicates and individuals of your choice.
Lines that are contained within the same number belong to one formula.
These examples purposely contain some expressions of natural language that are not so easily translatable
into predicate logic. Try to think of what the sentences mean from a narrow semantical point of view
and reduce it to a fully truth-conditional semantics, filtering out word material or additional meaning
that has no equivalent in our logic.
a) (1)
(2)
(3)
(4)
Some roses are red
All violets are blue
Jim is a pink hat
Noone reads PTQ
b) (5) All roses are bad
and most violets are too
(6) If you don’t love me
then I don’t love you
c) (7) If violets are red
and roses are blue
then every student
loves Montague
Solution (partial):
(1) some
= λP λQ∃x[P (x) ∧ Q(x)]
some roses
= λP λQ∃x[P (x) ∧ Q(x)](rose′ )
= λQ∃x[rose′ (x) ∧ Q(x)]
some roses are red
= λQ∃x[rose′ (x) ∧ Q(x)](red′ )
= ∃x[rose′ (x) ∧ red′ (x)]
(2) all
= λP λQ∀x[P (x) → Q(x)]
all violets
= λP λQ∀x[P (x) → Q(x)](violet′ )
= λQ∀x[violet′ (x) → Q(x)]
all violets are blue
= λQ∀x[violet′ (x) → Q(x)](blue′ )
= ∀x[violet′ (x) → blue′ (x)]
(3) pink hat
= λx[pink ′ (x) ∧ hat′ (x)]
Jim is a pink hat
= λx[pink ′ (x) ∧ hat′ (x)](j ′ )
= [pink ′ (j ′ ) ∧ hat′ (j ′ )]
(4) reads
= λyλs[read′ (x, y)]
reads PTQ
– 3/5 –
= λyλx[read′ (x, y)](ptq ′ )
= λx[read′ (x, ptq ′ )]
no
= λP λQ¬∃x[P (x) ∧ Q(x)]
noone = no person
= λP λQ¬∃x[P (x) ∧ Q(x)](person′ )
= λQ¬∃x[person′ (x) ∧ Q(x)]
noone reads PTQ
= λQ¬∃x[person′ (x) ∧ Q(x)](λx[read′ (x, ptq ′ )])
= ¬∃x[person(x) ∧ λx[read(x, ptq ′ )]]
(5) most
= λQλP [Cardinality(λx[P (x) ∧ Q(x)]) > Cardinality(λx[P (x) ∧ ¬Q(x)])]
most violets
= λQλP [Cardinality(λx[P (x) ∧ Q(x)]) > Cardinality(λx[P (x) ∧ ¬Q(x)])](violet′ )
= λP [Cardinality(λx[violet′ (x) ∧ Q(x)]) > Cardinality(λx[violet′ (x) ∧ ¬Q(x)])]
most violets are bad (too)
= λP [Cardinality(λx[violet′ (x) ∧ Q(x)]) > Cardinality(λx[violet′ (x) ∧ ¬Q(x)])](bad′ )
= Cardinality(λx[violet′ (x) ∧ bad′ (x)]) > Cardinality(λx[violet′ (x) ∧ ¬bad′ (x)])
(6) love
= λyλx[love′ (x, y)]
love me
= λyλx[love′ (x, y)](i′ )
= λx[love′ (x, i′ )]
don’t
= λP λx[¬P (x)]
don’t love me
= λP λx[¬P (x)](λx[love′ (x, i′ )])
= λx[¬love′ (x, i′ )]
S1: You don’t love me
= λx[¬love′ (x, i′ )](y ′ )
= ¬love′ (y ′ , i′ )
love you
= λyλx[love′ (x, y)](y ′ )
= λx[love′ (x, y ′ )]
don’t
= λP λx[¬P (x)]
don’t love you
= λP λx[¬P (x)](λx[love′ (x, y ′ )])
= λx[¬love′ (x, y ′ )]
S2: I don’t love you
= λx[¬love′ (x, y ′ )](i′ )
= ¬love′ (i′ , y ′ )
If p, then q
= λpλq[p → q]
If you don’t love me, then q
= λpλq[p → q](S1)
= λq[¬love′ (y ′ , i′ ) → q]
If you don’t love me then I don’t love you
– 4/5 –
= λq[¬love′ (y ′ , i′ ) → q](S2)
= ¬love(y ′ , i′ ) → ¬love(i′ , y ′ )
(7) Depending on the interpretation of the DP ∅ roses (in general), either
(∀x[violet′ (x) → red′ (x)] ∧ ∀y[rose′ (y) → blue′ (y)]) → ∀z[student′ (z) → love(z, mont′ )] OR
(∃x[violet′ (x) ∧ red′ (x)] ∧ ∃y[rose′ (y) ∧ blue′ (y)]) → ∀z[student′ (z) → love(z, mont′ )]
– 5/5 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Wolfwalk Starfield
Tutoren: Thorwald Kassemek, Natalie Clarius
Exemplary Solutions:
Predicate Logic, Modal Logic,
and Intensional Type Theory
Zimmermann & Sternefeld: Ch. 10
Gamut: Vol. 2 Ch. 4-5
Partee, Meulen & Wall: Pt. D
Hodges: Ch. 43
Montague: PTQ
For a general instruction, see p.2.
Given is the following model:
M =< D, W, I >
D = {John, M ary, P aul}
W = {w1 , w2 , w3 }
I=
I(j)
I(m)
I(p)
I(student)
I(sleep)
w1
John
M ary
P aul
{John, M ary}
{John, P aul}
w2
John
M ary
P aul
{John, P aul}
{John, P aul}
w3
John
M ary
P aul
{M ary, P aul}
{John, M ary, P aul}
Evaluate the following expressions and determine their meaning in natural language:
a) ⟦∃x(student(x) ∧ sleep(x))⟧M,g,w1
b) ⟦◇student(j)⟧M,g,w2
c) ⟦◻sleep(m)⟧M,g,w3
d) ⟦∧ sleep⟧M,g,w2
e) ⟦∨∧ sleep⟧M,g,w2
f) ⟦∧ sleep(m)⟧M,g,w2
g) ⟦∨∧ sleep(m)⟧M,g,w2
h) ⟦◇∀x(student(x) → ¬student(x)⟧M,g,w3
i) ⟦◻∃x(student(x))⟧M,g,w1
(You can click on the formulas in the PDF file to jump directly to the corresponding solution.)
– 1/5 –
Solution:
The algorithm is always the same:
1. First look for the main operator. For simple statement and predicate logic this is the same as we
have done all the time, e.g. in (p ∨ q) ∧ r this would be the ∧; if there is one of the operators
∧ ∨
, , ◻, ◇, ∀, ∃ they will always have to be evaluated first going from left to right, unless of course
bracketing tells your something else.
For example in ◇ ◻ ∃x∀yφ the order of application would simply be ◇(◻∃...) then ◻(∃x∀...) then
∃x(∀...) then ∀y(φ), whereas in (◻φ) ∨ ψ you would have to first look at the ∨ and then at the ◻.
2. As soon as you have identified the main operator or connective, evaluate it by simply applying the
definitions from the book and replacing α by whatever follows the symbol in question; the syntax
and semantics statements unambiguously tell you how to evaluate a statement given the main
operator or connective. You now have to do no more than writing this definition down, filling the
arguments in and then mechanically applying what the evaluation rules tell you.
3. Apply this recursively over and over again until you reach the atomic parts from which you can
compositionally derive the extension of the whole expression.
4. You will frequently encounter steps where you have to check whether ”for all xyz it holds that...”
or ”there is a xyz such that...”, with ”xyz” being assignment functions or possible worlds. To
evaluate these, proceed as follows:
● If there is an existential quantifier (∃g or ∃w):
● If you immediately see that there is one particular g/w that makes the statement true, just
provide this one and you’re done.
● If there is none that makes it true, you have to go through every possible g/w to show that
all of them fail and therefore the outer statement is false.
● If there is a universal quantifier (∀g or ∀w):
● If you immediately see that there is one for which the inner statement does not hold, you only
have to provide this one to show that the outer statement is not true.
● If the inner statement is true of all g/w, you have to provide the proof for each of them before
confirming that the whole statement is true.
5. The rest is, as said, just recursive application of always the same definitions.
– 2/5 –
a) A student sleeps in w1 (true)
⟦∃x(student(x) ∧ sleep(x))⟧M,g,w1
= 1 iff there is an assignment function g ′ ∶ ⟦student(x) ∧ sleep(x)⟧M,g,w1 = 1
● For g ′ (x) = John ∶ ⟦(student(x) ∧ sleep(x)⟧M,g′ ,w1
= 1 iff ⟦student⟧M,g′ ,w1 (⟦x⟧M,g′ ,w1 ) = 1 and ⟦sleep⟧M,g′ ,w1 (⟦x⟧M,g′ ,w1 ) = 1
= 1 iff g ′ (x) ∈ I(student)(w1 ) and g ′ (x) ∈ I(sleep)(w1 )
= 1 iff John ∈ {John, M ary} and John ∈ {John, P aul}
= 1 since John ∈ {John, M ary} is true and John ∈ {John, P aul} is true
There is an assignment function g ′ such that ⟦student(x) ∧ sleep(x)⟧M,g′ ,w1 = 1 with g ′ (x) = John
Therefore, ⟦∃x(student(x) ∧ sleep(x))⟧M,g,w1 = 1
b) It is possible that John is a student (true)
⟦◇student(j)⟧M,g,w2
= 1 iff there is a world w′ ∈ W ∶ ⟦student(j)⟧M,g,w′ = 1
● For w′ = w1 ∶ ⟦student(j)⟧M,g,w1
= 1 iff ⟦student⟧M,g,w1 (⟦j⟧M,g,w1 )
= 1 iff I(j)(w1 ) ∈ I(student)(w1 )
= 1 iff John ∈ {John, M ary}
=1
There is a world w′ such that ⟦student(j)⟧M,g,w′ = 1 with w′ = w1
Therefore, ⟦◇student(j)⟧M,g,w2 = 1
c) It is necessary that Mary sleeps (false)
⟦◻sleep(m)⟧M,g,w3
= 1 iff for all worlds w′ ∈ W ∶ ⟦sleep(m)⟧M,g,w′ = 1
● For w′ = w1 : ⟦sleep(m)⟧M,g,w1
= 1 iff ⟦sleep⟧M,g,w1 (⟦m⟧M,g,w1 )
= 1 iff I(m)(w1 ) ∈ I(sleep)(w1 )
= 1 iff M ary ∈ {John, P aul}
=0
There is an world w′ such that it does not hold that ⟦student(j)⟧M,g,w′ = 1 with w′ = w1
Therefore, not for all worlds w′ ∈ W it holds that ⟦sleep(m)⟧M,g,w′ = 1
Therefore, ⟦◻sleep(m)⟧M,g,w3 = 0
d) (The concept of sleep in all possible worlds)
⟦∧ sleep⟧M,g,w2
= that function h ∈ DW such that for all w′ ∈ W ∶ h(w′ ) = ⟦sleep⟧M,g,w′
= I(sleep)
● For w′ = w1 ∶ h(w1 ) = ⟦sleep⟧M,g,w1 = I(sleep)(w1 ) = {John, P aul}
● For w′ = w2 ∶ h(w2 ) = ⟦sleep⟧M,g,w2 = I(sleep)(w2 ) = {John, P aul}
● For w′ = w3 ∶ h(w3 ) = ⟦sleep⟧M,g,w3 = I(sleep)(w3 ) = {John, P aul, M ary}
e) (The set of sleepers in w2 )
⟦∨∧ sleep⟧M,g,w2
= ⟦sleep⟧M,g,w2
= I(sleep)(w2 )
– 3/5 –
= {John, P aul}
f) (The set of possible worlds in which Mary sleeps)
⟦∧ sleep(m)⟧M,g,w2
= that function h ∈ DW such that for all w′ ∈ W ∶ h(w′ ) = ⟦sleep(m)⟧M,g,w′
● For w′ = w1 ∶ h(w1 ) = ⟦sleep(m)⟧w1 = 0
● For w′ = w2 ∶ h(w2 ) = ⟦sleep(m)⟧w2 = 0
● For w′ = w3 ∶ h(w3 ) = ⟦sleep(m)⟧w3 = 1
g) Mary sleeps in w2 (false)
⟦∨∧ sleep(m)⟧M,g,w2
= ⟦sleep(m)⟧M,g,w2
= ⟦sleep⟧M,w,w2 (⟦m⟧M,g,w2 )
= 1 iff I(m)(w2 ) ∈ I(sleep)(w2 )
= 1 iff M ary ∈ {John, P aul}
=0
h) It is possible that every student is not a student (false)
⟦◇∀x(student(x) → ¬student(x)⟧M,g,w3
= 1 iff there is a world w′ ∈ W ∶ ⟦∀x(student(x) → ¬student(x)⟧M,g,w′ = 1
Here you may abbreviate a little: A → ¬A can only be true iff A is false. This would mean that
there must be no students in that world. Since this is in no world the case, the whole statement
must be false.
Detailed solution:
● For w′ = w1 ∶ ⟦∀x(student(x) → ¬student(x)⟧M,g,w1
= 1 iff for all assignment functions g ′ ∶ ⟦student(x) → ¬student(x)⟧M,g′ ,w1 = 1
● For g ′ (x) = John: false
● For g ′ (x) = M ary: false
● For g ′ (x) = P aul: false
There is an assignment function g ′ such that it does not hold that ⟦student(x) → ¬student(x)⟧M,g′ ,w1 =
1 with g ′ (x) = John (or g ′ (x) = M ary or g ′ (x) = P aul)
Therefore, not for all assignment functions g ′ it holds that ⟦student(x) → ¬student(x)⟧M,g′ ,w1 =
1
Therefore, ⟦∀x(student(x) → ¬student(x)⟧M,g,w1 = 0
● For w′ = w2 : (similar as above; ⟦∀x(student(x) → ¬student(x)⟧M,g,w2 = 0)
● For w′ = w2 : (similar as above; ⟦∀x(student(x) → ¬student(x)⟧M,g,w3 = 0)
There is no world w′ ∈ W such that ⟦∀x(student(x) → ¬student(x)⟧M,g,w′ = 1
Therefore, ⟦◇∀x(student(x) → ¬student(x)⟧M,g,w3 = 0
i) It is necessary that there is a student (true)
⟦◻∃x(student(x))⟧M,g,w1
= 1 iff for all worlds w′ ∈ W ∶ ⟦∃x(student(x))⟧M,g,w′ = 1
Here you may abbreviate again: You can already informally infer from h) that this must be true.
Detailed solution:
● For w′ = w1 ∶ ⟦∃x(student(x))⟧M,g,w1
= 1 iff there is an assignment function g ′ ∶ ⟦student(x)⟧M,g′ ,w1 = 1
– 4/5 –
● For g ′ (x) = M ary⟦student(x)⟧M,g′ ,w1
= ⟦student⟧M,g′ ,w1 (⟦x⟧M,g′ ,w1 )
= 1 iff g ′ (x) ∈ I(student(w1 ))
= 1 iff M ary ∈ {John, M ary}
=1
There is an assignment function g ′ such that ⟦student(x)⟧M,g′ ,w1 = 1 with g ′ (x) = M ary
(works also with g ′ (x) = John)
Therefore, ⟦∃x(student(x))⟧M,g,w1 = 1
● For w′ = w2 : (same as above, with g ′ (x) = John or g ′ (x) = P aul)
● For w′ = w3 : (same as above, with g ′ (x) = M ary or g ′ (x) = P aul)
Therefore, in all worlds w′ ∈ W it holds that ⟦∃x(student(x))⟧M,g,w′ = 1
Therefore, ⟦◻∃x(student(x))⟧M,g,w1 = 1
– 5/5 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Feldgang Sternewolf
Tutoren: Thorwald Kassemek, Natalie Clarius
Mock exam:
Midterm
Zimmermann & Sternefeld: Ch. 1-8
Slides ESSLLI: No. 1-5 + Addenda
Gamut: Vol. 1 Ch. 1-4
Partee, Meulen & Wall: Pt. A-B
Hodges: Ch. 1-41
Exercise 1: Prerequisites
1. Explain the difference between direct and indirect semantic interpretation.
2. Determine which of the following forumale are equivalent:
(a) (A ∧ ¬(A ∧ ¬B))
(b) ((A → B) ∧ A)
(c) (¬(¬A ∨ (A → B)))
3. Give a predicate logic translation of the sentence
For every entity it holds, if it is an animal and pink, then it is a panther.
using constant symbols of your choice.
4. What does the ”most certain principle” state and who does it attribute to?
Exercise 2: Extensions
1. Determine the class (i.e. functional noun, count noun, verb, ...) as well as the type of the extension
(i.e. whether it is an individual, a truth value, ...)of the following expressions:
(a) house
(b) brother of
(c) sing
(d) milk
(e) donate
(f) large
(g) mayor of
(h) hit
2. State four ways of composing extensions, and give one example for each.
3. What does plugging of the verb love Mary with the indiviual John result in, i.e. what is its
extension and how the operation affect the properties of the verb?
4. Explain how the extension of the table is compsed of its immediate parts, i.e. what the semantics
of the individual constituents are and what happens during their combination.
– 1/2 –
5. What is the required operation for composing the extension of red apple, and why does it cause
problems for the syntactically similar term alleged murderer ?
6. How does quantifier raising affect the extension of the declarative sentence the QDP has bee moved
out, and what is the term for this kind of operation?
Exercise 3: Determiners and quantifiers
Consider the following sentence:
A single falls in love with every woman on Parhsip.1
1. Determine the two readings and explain how the ambiguity arises. Which reading is the desired
one, and what kind of syntactic operation is needed for deriving the other one?
2. Give the logical form of each reading, using either boxes, brackets or a tree, and additionaly a
corresponding predicate logic formula.
3. Calculate the truth conditions of the two readings using two different methods, and explain how
those methods correspdond to the results.
4. In what way can bound variable pronouns be accounted for, and how is this related to an elementary
problem about generalised quantifiers?
5. Give a definition for each of the following expressions, and explain why one of them has a special
status:
(a) every
(b) most
(c) some
(d) a
(e) the
(f) no
(g) nothing
6. Explain the paradox of composing the extension of the + plural-NP.
7. Give the extension of three different readings for the sentence The boys and the girls met in the
cinema.
8. Give one solution for a tree of the logical types being composed of the immediate constituents in
the slightly changed sentence:
The single loves every woman on Parship.
Exercise 4: Propositions and intensions
Consider the sentece
John knows that grass is green.
1. Explain the terms logical space, proposition and possible worlds with respect to each other.
2. What is the extension of the above sentence as a whole, and what is the proposition of the embedded
sentence?
3. What is the intension of an expression in general and what is the intension of the sentence above
in particular?
1 The source and original version of this sentence is the (in Germany) well-known advertising Alle elf Minuten verliebt
sich ein Single über Parhsip, which is a nice example for scope ambiguity in real life - but since we can not porperly analyse
”alle elf Minuten” with our methods, we had to slightly adapt the example.
– 2/2 –
Eberhard Karls Universität Tübingen | Seminar für Sprachwissenschaft (SfS)
PS Semantik 1 (WS 2015/16), Prof. Dr. Wolfgang Sternefeld
Tutoren: Thorwald Kassemek, Natalie Clarius
Keywords:
Final Exam
Zimmermann & Sternefeld: Ch. 10
Gamut: Vol. 1 Ch. 1-4
Gamut: Vol. 2 Ch. 4-5
Partee, Meulen & Wall: Pt. A-B, D
Levinson: Ch. 4
Montague: PTQ
This is a list of some important keywords for the final exam that you should be familiar with.
And no, of course it is not sufficient to just learn the definitions by heart, you should also know what
they actually mean and how to semantics ;)
A) presuppositions
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
definite article, existence and uniqueness condition, naive vs. quanitifactional analysis
entailment relations
veridical vs. factive verbs
presupposition and assertion
presupposition and discourse, common ground (CG), context update, informativity
presupposition accomodation
presupposition projection
B) predicate logic
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
scope ambiguity and LF
assignment functions, x-alternatives
models
logical types, intensional type theory
lambda-calculus, functional application and compositionality
possible worlds, ◻ and ◇
intensionality, ∧ and ∨
Montague’s PTQ
– 1/1 –